Question

The cylinder in the figure(Figure 1) has a moveable piston attached to a spring. The cylinder's cross-section area is 10 cm2, it contains 0.0040mol of gas, and the spring constant is 1500 N/m. At 19 degreesC the spring is neither compressed nor stretched.

How far is the spring compressed if the gas temperature is raised to 200 degrees C ?

Answer 1

The initial pressure of the gas P1 = 1 atm

The initial volume of the gas is V1 = AL₀

Where A is the cross-sectional area = 10 cm^2 = 10^-3 m^2

L₀ is the uncompressed length of the spring

The initial temperature of the gas is T1 = 21 C = 294 K

The number of moles of gas is n = 0.0044 mol

The spring constant is k = 1500 N/m

The final temperture of the gas is T2 = 130 C = 403 K

Let ?x be the change in length of the spring,

Then the final volume become V2 = A(L₀ +?x)

the increse in pressure will be

P2 = P1 + F/A

P2 = P1 + k?x/A (Hook's law)

By the ideal gas law

P1V1/T1 = P2V2/T2

P1(AL₀)/T1 = (P1 + k?x/A)[A(L₀ +?x)]/T2

T2/T1 = [1 + k?x/P1A][1 +?x/L₀]

Also

P1V1 = nRT1

P1(AL₀) = nRT1

L₀ = 0.0044*8.314*294/1*10^-3

L₀ = 0.01075 m

403/294 = [1 + 1500*?x/1*10^-3][1 +?x/0.01075]

1.371 = (1+1.5?x)(1+93.023?x)

139.5345?x^2 + 94.523?x - 0.371 = 0

Solving the above equation,

?x = 0.0039 m = 3.9 mm

Therefore the spring is compressed by 3.9 mm